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Given positive numbers N, K, D (1<= N <= 10^5, 1<=K<=N, 1<=D<=9). How many numbers with N digits are there, that have K consecutive digits D? Write the answer mod (10^9 + 7).
For example: N = 4, K = 3, D = 6, there are 18 numbers:
1666, 2666, 3666, 4666, 5666, 6660,
6661, 6662, 6663, 6664, 6665, 6666, 6667, 6668, 6669, 7666, 8666 and 9666.
Can we calculate the answer in O(N*K) (maybe dynamic programming)?
I've tried using combination.
If
N = 4, K = 3, D = 6. The number I have to find is abcd.
+) if (a = b = c = D), I choose digit for d. There are 10 ways (6660, 6661, 6662, 6663, 6664, 6665, 6666, 6667, 6668, 6669)
+) if (b = c = d = D), I choose digit for a (a > 0). There are 9 ways (1666, 2666, 3666, 4666, 5666, 6666, 7666, 8666, 9666)
But in two cases, the number 6666 is counted twice. N and K is very large, how can I count all of them?
If one is looking for a mathematical solution (vs. necessarily an algorithmic one) it's good to look at it in terms of the base cases and some formulas. They might turn out to be something you can do some kind of refactoring and get a tidy formula for. So just for the heck of it...here's a take on it that doesn't deal with the special treatment of zeros. Because that throws some wrenches in.
Let's look at a couple of base cases, and call our answer F(N,K) (not considering D, as it isn't relevant to account for; but taking it as a parameter anyway.):
when N = 0
You'll never find any length sequences of digits when there's no digit.
F(0, K) = 0 for any K.
when N = 1
Fairly obvious. If you're looking for K sequential digits in a single digit, the options are limited. Looking for more than one? No dice.
F(1, K) = 0 for any K > 1
Looking for exactly one? Okay, there's one.
F(1, 1) = 1
Sequences of zero sequential digits allowed? Then all ten digits are fine.
F(1, 0) = 10
for N > 1
when K = 0
Basically, all N-digit numbers will qualify. So the number of possibilities meeting the bar is 10^N. (e.g. when N is 3 then 000, 001, 002, ... 999 for any D)
F(N, 0) = 10^N for any N > 1
when K = 1
Possibilities meeting the condition is any number with at least one D in it. How many N-digit numbers are there which contain at least one digit D? Well, it's going to be 10^N minus all the numbers that have no instances of the digit D. 10^N - 9^N
F(N, 1) = 10^N - 9^N for any N > 1
when N < K
No way to get K sequential digits if N is less than K
F(N, K) = 0 when N < K
when N = K
Only one possible way to get K sequential digits in N digits.
F(N, K) = 1 when N = K
when N > K
Okay, we already know that N > 1 and K > 1. So this is going to be the workhorse where we hope to use subexpressions for things we've already solved.
Let's start by considering popping off the digit at the head, leaving N-1 digits on the tail. If that N-1 series could achieve a series of K digits all by itself, then adding another digit will not change anything about that. That gives us a term 10 * F(N-1, K)
But if our head digit is a D, that is special. Our cases will be:
It might be the missing key for a series that started with K-1 instances of D, creating a full range of K.
It might complete a range of K-1 instances of D, but on a case that already had a K series of adjacent D (that we thus accounted for in the above term)
It might not help at all.
So let's consider two separate categories of tail series: those that start with K-1 instances of D and those that do not. Let's say we have N=7 shown as D:DDDXYZ and with K=4. We subtract one from N and from K to get 6 and 3, and if we subtract them we get how many trailing any-digits (XYZ here) are allowed to vary. Our term for the union of (1) and (2) to add in is 10^((N-1)-(K-1)).
Now it's time for some subtraction for our double-counts. We haven't double counted any cases that didn't start with K-1 instances of D, so we keep our attention on that (DDDXYZ). If the value in the X slot is a D then it was definitely double counted. We can subtract out the term for that as 10^(((N - 1) - 1) - (K - 1)); in this case giving us all the pairings of YZ digits you can get with X as D. (100).
The last thing to get rid of are the cases where X is not a D, but in whatever that leftover after the X position there was still a K length series of D. Again we reuse our function, and subtract a term 9 * F(N - K, K, D).
Paste it all together and simplify a couple of terms you get:
F(N, K) = 10 * F(N-1,K,D) + 10^(N-K) - 10^(10,N-K-1) - 9 * F(N-K-1,K,D)
Now we have a nice functional definition suitable for Haskelling or whatever. But I'm still awkward with that, and it's easy enough to test in C++. So here it is (assuming availability of a long integer power function):
long F(int N, int K, int D) {
if (N == 0) return 0;
if (K > N) return 0;
if (K == N) return 1;
if (N == 1) {
if (K == 0) return 10;
if (K == 1) return 1;
return 0;
}
if (K == 0)
return power(10, N);
if (K == 1)
return power(10, N) - power(9, N);
return (
10 * F(N - 1, K, D)
+ power(10, N - K)
- power(10, N - K - 1)
- 9 * F(N - K - 1, K, D)
);
}
To double-check this against an exhaustive generator, here's a little C++ test program that builds the list of vectors that it scans using std::search_n. It checks the slow way against the fast way for N and K. I ran it from 0 to 9 for each:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
// http://stackoverflow.com/a/1505791/211160
long power(int x, int p) {
if (p == 0) return 1;
if (p == 1) return x;
long tmp = power(x, p/2);
if (p%2 == 0) return tmp * tmp;
else return x * tmp * tmp;
}
long F(int N, int K, int D) {
if (N == 0) return 0;
if (K > N) return 0;
if (K == N) return 1;
if (N == 1) {
if (K == 0) return 10;
if (K == 1) return 1;
return 0;
}
if (K == 0)
return power(10, N);
if (K == 1)
return power(10, N) - power(9, N);
return (
10 * F(N - 1, K, D)
+ power(10, N - K)
- power(10, N - K - 1)
- 9 * F(N - K - 1, K, D)
);
}
long FSlowCore(int N, int K, int D, vector<int> & digits) {
if (N == 0) {
if (search_n(digits.begin(), digits.end(), K, D) != end(digits)) {
return 1;
} else
return 0;
}
long total = 0;
digits.push_back(0);
for (int curDigit = 0; curDigit <= 9; curDigit++) {
total += FSlowCore(N - 1, K, D, digits);
digits.back()++;
}
digits.pop_back();
return total;
}
long FSlow(int N, int K, int D) {
vector<int> digits;
return FSlowCore(N, K, D, digits);
}
bool TestF(int N, int K, int D) {
long slow = FSlow(N, K, D);
long fast = F(N, K, D);
cout << "when N = " << N
<< " and K = " << K
<< " and D = " << D << ":\n";
cout << "Fast way gives " << fast << "\n";
cout << "Slow way gives " << slow << "\n";
cout << "\n";
return slow == fast;
}
int main() {
for (int N = 0; N < 10; N++) {
for (int K = 0; K < 10; K++) {
if (!TestF(N, K, 6)) {
exit(1);
}
}
}
}
Of course, since it counts leading zeros it will be different from the answers you got. See the test output here in this gist.
Modifying to account for the special-case zero handling is left as an exercise for the reader (as is modular arithmetic). Eliminating the zeros make it messier. Either way, this may be an avenue of attack for reducing the number of math operations even further with some transformations...perhaps.
Miquel is almost correct, but he missed a lot of cases. So, with N = 8, K = 5, and D = 6, we will need to look for those numbers that has the form:
66666xxx
y66666xx
xy66666x
xxy66666
with additional condition that y cannot be D.
So we can have our formula for this example:
66666xxx = 10^3
y66666xx = 8*10^2 // As 0 can also not be the first number
xy66666x = 9*9*10
xxy66666 = 9*10*9
So, the result is 3420.
For case N = 4, K = 3 and D = 6, we have
666x = 10
y666 = 8//Again, 0 is not counted!
So, we have 18 cases!
Note: We need to be careful that the first number cannot be 0! And we need to handle the case when D is zero too!
Update Java working code, Time complexity O(N-K)
static long cal(int n, int k, int d) {
long Mod = 1000000007;
long result = 0;
for (int i = 0; i <= n - k; i++) {//For all starting positions
if (i != 0 || d != 0) {
int left = n - k;
int upper_half = i;//Amount of digit that preceding DDD
int lower_half = n - k - i;//Amount of digit following DDD
long tmp = 1;
if (upper_half == 1) {
if (d == 0) {
tmp = 9;
} else {
tmp = 8;
}
}else if(upper_half >= 2){
//The pattern is x..yDDD...
tmp = (long) (9 * 9 * Math.pow(10, upper_half - 2));
}
tmp *= Math.pow(10, lower_half);
//System.out.println(tmp + " " + upper_half + " " + lower_half);
result += tmp;
result %= Mod;
}
}
return result;
}
Sample Tests:
N = 8, K = 5, D = 6
Output
3420
N = 4, K = 3, D = 6
Output
18
N = 4, K = 3, D = 0
Output
9
I need to calculate combinations for a number.
What is the fastest way to calculate nCp where n>>p?
I need a fast way to generate binomial coefficients for an polynomial equation and I need to get the coefficient of all the terms and store it in an array.
(a+b)^n = a^n + nC1 a^(n-1) * b + nC2 a^(n-2) * ............
+nC(n-1) a * b^(n-1) + b^n
What is the most efficient way to calculate nCp ??
You cau use dynamic programming in order to generate binomial coefficients
You can create an array and than use O(N^2) loop to fill it
C[n, k] = C[n-1, k-1] + C[n-1, k];
where
C[1, 1] = C[n, n] = 1
After that in your program you can get the C(n, k) value just looking at your 2D array at [n, k] indices
UPDATE smth like that
for (int k = 1; k <= K; k++) C[0][k] = 0;
for (int n = 0; n <= N; n++) C[n][0] = 1;
for (int n = 1; n <= N; n++)
for (int k = 1; k <= K; k++)
C[n][k] = C[n-1][k-1] + C[n-1][k];
where the N, K - maximum values of your n, k
If you need to compute them for all n, Ribtoks's answer is probably the best.
For a single n, you're better off doing like this:
C[0] = 1
for (int k = 0; k < n; ++ k)
C[k+1] = (C[k] * (n-k)) / (k+1)
The division is exact, if done after the multiplication.
And beware of overflowing with C[k] * (n-k) : use large enough integers.
If you want complete expansions for large values of n, FFT convolution might be the fastest way. In the case of a binomial expansion with equal coefficients (e.g. a series of fair coin tosses) and an even order (e.g. number of tosses) you can exploit symmetries thus:
Theory
Represent the results of two coin tosses (e.g. half the difference between the total number of heads and tails) with the expression A + A*cos(Pi*n/N). N is the number of samples in your buffer - a binomial expansion of even order O will have O+1 coefficients and require a buffer of N >= O/2 + 1 samples - n is the sample number being generated, and A is a scale factor that will usually be either 2 (for generating binomial coefficients) or 0.5 (for generating a binomial probability distribution).
Notice that, in frequency, this expression resembles the binomial distribution of those two coin tosses - there are three symmetrical spikes at positions corresponding to the number (heads-tails)/2. Since modelling the overall probability distribution of independent events requires convolving their distributions, we want to convolve our expression in the frequency domain, which is equivalent to multiplication in the time domain.
In other words, by raising our cosine expression for the result of two tosses to a power (e.g. to simulate 500 tosses, raise it to the power of 250 since it already represents a pair), we can arrange for the binomial distribution for a large number to appear in the frequency domain. Since this is all real and even, we can substitute the DCT-I for the DFT to improve efficiency.
Algorithm
decide on a buffer size, N, that is at least O/2 + 1 and can be conveniently DCTed
initialise it with the expression pow(A + A*cos(Pi*n/N),O/2)
apply the forward DCT-I
read out the coefficients from the buffer - the first number is the central peak where heads=tails, and subsequent entries correspond to symmetrical pairs successively further from the centre
Accuracy
There's a limit to how high O can be before accumulated floating-point rounding errors rob you of accurate integer values for the coefficients, but I'd guess the number is pretty high. Double-precision floating-point can represent 53-bit integers with complete accuracy, and I'm going to ignore the rounding loss involved in the use of pow() because the generating expression will take place in FP registers, giving us an extra 11 bits of mantissa to absorb the rounding error on Intel platforms. So assuming we use a 1024-point DCT-I implemented via the FFT, that means losing 10 bits' accuracy to rounding error during the transform and not much else, leaving us with ~43 bits of clean representation. I don't know what order of binomial expansion generates coefficients of that size, but I dare say it's big enough for your needs.
Asymmetrical expansions
If you want the asymmetrical expansions for unequal coefficients of a and b, you'll need to use a two-sided (complex) DFT and a complex pow() function. Generate the expression A*A*e^(-Pi*i*n/N) + A*B + B*B*e^(+Pi*i*n/N) [using the complex pow() function to raise it to the power of half the expansion order] and DFT it. What you have in the buffer is, again, the central point (but not the maximum if A and B are very different) at offset zero, and it is followed by the upper half of the distribution. The upper half of the buffer will contain the lower half of the distribution, corresponding to heads-minus-tails values that are negative.
Notice that the source data is Hermitian symmetrical (the second half of the input buffer is the complex conjugate of the first), so this algorithm is not optimal and can be performed using a complex-to-complex FFT of half the required size for optimum efficiency.
Needless to say, all the complex exponentiation will chew more CPU time and hurt accuracy compared to the purely real algorithm for symmetrical distributions above.
This is my version:
def binomial(n, k):
if k == 0:
return 1
elif 2*k > n:
return binomial(n,n-k)
else:
e = n-k+1
for i in range(2,k+1):
e *= (n-k+i)
e /= i
return e
I recently wrote a piece of code that needed to call for a binary coefficient about 10 million times. So I did a combination lookup-table/calculation approach that's still not too wasteful of memory. You might find it useful (and my code is in the public domain). The code is at
http://www.etceterology.com/fast-binomial-coefficients
It's been suggested that I inline the code here. A big honking lookup table seems like a waste, so here's the final function, and a Python script that generates the table:
extern long long bctable[]; /* See below */
long long binomial(int n, int k) {
int i;
long long b;
assert(n >= 0 && k >= 0);
if (0 == k || n == k) return 1LL;
if (k > n) return 0LL;
if (k > (n - k)) k = n - k;
if (1 == k) return (long long)n;
if (n <= 54 && k <= 54) {
return bctable[(((n - 3) * (n - 3)) >> 2) + (k - 2)];
}
/* Last resort: actually calculate */
b = 1LL;
for (i = 1; i <= k; ++i) {
b *= (n - (k - i));
if (b < 0) return -1LL; /* Overflow */
b /= i;
}
return b;
}
#!/usr/bin/env python3
import sys
class App(object):
def __init__(self, max):
self.table = [[0 for k in range(max + 1)] for n in range(max + 1)]
self.max = max
def build(self):
for n in range(self.max + 1):
for k in range(self.max + 1):
if k == 0: b = 1
elif k > n: b = 0
elif k == n: b = 1
elif k == 1: b = n
elif k > n-k: b = self.table[n][n-k]
else:
b = self.table[n-1][k] + self.table[n-1][k-1]
self.table[n][k] = b
def output(self, val):
if val > 2**63: val = -1
text = " {0}LL,".format(val)
if self.column + len(text) > 76:
print("\n ", end = "")
self.column = 3
print(text, end = "")
self.column += len(text)
def dump(self):
count = 0
print("long long bctable[] = {", end="");
self.column = 999
for n in range(self.max + 1):
for k in range(self.max + 1):
if n < 4 or k < 2 or k > n-k:
continue
self.output(self.table[n][k])
count += 1
print("\n}}; /* {0} Entries */".format(count));
def run(self):
self.build()
self.dump()
return 0
def main(args):
return App(54).run()
if __name__ == "__main__":
sys.exit(main(sys.argv))
If you really only need the case where n is much larger than p, one way to go would be to use the Stirling's formula for the factorials. (if n>>1 and p is order one, Stirling approximate n! and (n-p)!, keep p! as it is etc.)
The fastest reasonable approximation in my own benchmarking is the approximation used by the Apache Commons Maths library: http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/special/Gamma.html#logGamma(double)
My colleagues and I tried to see if we could beat it, while using exact calculations rather than approximates. All approaches failed miserably (many orders slower) except one, which was 2-3 times slower. The best performing approach uses https://math.stackexchange.com/a/202559/123948, here is the code (in Scala):
var i: Int = 0
var binCoeff: Double = 1
while (i < k) {
binCoeff *= (n - i) / (k - i).toDouble
i += 1
}
binCoeff
The really bad approaches where various attempts at implementing Pascal's Triangle using tail recursion.
nCp = n! / ( p! (n-p)! ) =
( n * (n-1) * (n-2) * ... * (n - p) * (n - p - 1) * ... * 1 ) /
( p * (p-1) * ... * 1 * (n - p) * (n - p - 1) * ... * 1 )
If we prune the same terms of the numerator and the denominator, we are left with minimal multiplication required. We can write a function in C to perform 2p multiplications and 1 division to get nCp:
int binom ( int p, int n ) {
if ( p == 0 ) return 1;
int num = n;
int den = p;
while ( p > 1 ) {
p--;
num *= n - p;
den *= p;
}
return num / den;
}
I was looking for the same thing and couldn't find it, so wrote one myself that seems optimal for any Binomial Coeffcient for which the endresult fits into a Long.
// Calculate Binomial Coefficient
// Jeroen B.P. Vuurens
public static long binomialCoefficient(int n, int k) {
// take the lowest possible k to reduce computing using: n over k = n over (n-k)
k = java.lang.Math.min( k, n - k );
// holds the high number: fi. (1000 over 990) holds 991..1000
long highnumber[] = new long[k];
for (int i = 0; i < k; i++)
highnumber[i] = n - i; // the high number first order is important
// holds the dividers: fi. (1000 over 990) holds 2..10
int dividers[] = new int[k - 1];
for (int i = 0; i < k - 1; i++)
dividers[i] = k - i;
// for every dividers there is always exists a highnumber that can be divided by
// this, the number of highnumbers being a sequence that equals the number of
// dividers. Thus, the only trick needed is to divide in reverse order, so
// divide the highest divider first trying it on the highest highnumber first.
// That way you do not need to do any tricks with primes.
for (int divider: dividers) {
boolean eliminated = false;
for (int i = 0; i < k; i++) {
if (highnumber[i] % divider == 0) {
highnumber[i] /= divider;
eliminated = true;
break;
}
}
if(!eliminated) throw new Error(n+","+k+" divider="+divider);
}
// multiply remainder of highnumbers
long result = 1;
for (long high : highnumber)
result *= high;
return result;
}
If I understand the notation in the question, you don't just want nCp, you actually want all of nC1, nC2, ... nC(n-1). If this is correct, we can leverage the following relationship to make this fairly trivial:
for all k>0: nCk = prod_{from i=1..k}( (n-i+1)/i )
i.e. for all k>0: nCk = nC(k-1) * (n-k+1) / k
Here's a python snippet implementing this approach:
def binomial_coef_seq(n, k):
"""Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""
b = [1]
for i in range(1,k+1):
b.append(b[-1] * (n-i+1)/i)
return b
If you need all coefficients up to some k > ceiling(n/2), you can use symmetry to reduce the number of operations you need to perform by stopping at the coefficient for ceiling(n/2) and then just backfilling as far as you need.
import numpy as np
def binomial_coef_seq2(n, k):
"""Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""
k2 = int(np.ceiling(n/2))
use_symmetry = k > k2
if use_symmetry:
k = k2
b = [1]
for i in range(1, k+1):
b.append(b[-1] * (n-i+1)/i)
if use_symmetry:
v = k2 - (n-k)
b2 = b[-v:]
b.extend(b2)
return b
Time Complexity : O(denominator)
Space Complexity : O(1)
public class binomialCoeff {
static double binomialcoeff(int numerator, int denominator)
{
double res = 1;
//invalid numbers
if (denominator>numerator || denominator<0 || numerator<0) {
res = -1;
return res;}
//default values
if(denominator==numerator || denominator==0 || numerator==0)
return res;
// Since C(n, k) = C(n, n-k)
if ( denominator > (numerator - denominator) )
denominator = numerator - denominator;
// Calculate value of [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
while (denominator>=1)
{
res *= numerator;
res = res / denominator;
denominator--;
numerator--;
}
return res;
}
/* Driver program to test above function*/
public static void main(String[] args)
{
int numerator = 120;
int denominator = 20;
System.out.println("Value of C("+ numerator + ", " + denominator+ ") "
+ "is" + " "+ binomialcoeff(numerator, denominator));
}
}
This is an interview question: "Given 2 integers x and y, check if x is an integer power of y" (e.g. for x = 8 and y = 2 the answer is "true", and for x = 10 and y = 2 "false").
The obvious solution is:int n = y; while(n < x) n *= y; return n == x
Now I am thinking about how to improve it.
Of course, I can check some special cases: e.g. both x and y should be either odd or even numbers, i.e. we can check the least significant bit of x and y. However I wonder if I can improve the core algorithm itself.
You'd do better to repeatedly divide y into x. The first time you get a non-zero remainder you know x is not an integer power of y.
while (x%y == 0) x = x / y
return x == 1
This deals with your odd/even point on the first iteration.
It means logy(x) should be an integer. Don't need any loop. in O(1) time
public class PowerTest {
public static boolean isPower(int x, int y) {
double d = Math.log(Math.abs(x)) / Math.log(Math.abs(y));
if ((x > 0 && y > 0) || (x < 0 && y < 0)) {
if (d == (int) d) {
return true;
} else {
return false;
}
} else if (x > 0 && y < 0) {
if ((int) d % 2 == 0) {
return true;
} else {
return false;
}
} else {
return false;
}
}
/**
* #param args
*/
public static void main(String[] args) {
System.out.println(isPower(-32, -2));
System.out.println(isPower(2, 8));
System.out.println(isPower(8, 12));
System.out.println(isPower(9, 9));
System.out.println(isPower(-16, 2));
System.out.println(isPower(-8, -2));
System.out.println(isPower(16, -2));
System.out.println(isPower(8, -2));
}
}
This looks for the exponent in O(log N) steps:
#define MAX_POWERS 100
int is_power(unsigned long x, unsigned long y) {
int i;
unsigned long powers[MAX_POWERS];
unsigned long last;
last = powers[0] = y;
for (i = 1; last < x; i++) {
last *= last; // note that last * last can overflow here!
powers[i] = last;
}
while (x >= y) {
unsigned long top = powers[--i];
if (x >= top) {
unsigned long x1 = x / top;
if (x1 * top != x) return 0;
x = x1;
}
}
return (x == 1);
}
Negative numbers are not handled by this code, but it can be done easyly with some conditional code when i = 1
This looks to be pretty fast for positive numbers as it finds the lower and upper limits for desired power and then applies binary search.
#include <iostream>
#include <cmath>
using namespace std;
//x is the dividend, y the divisor.
bool isIntegerPower(int x, int y)
{
int low = 0, high;
int exp = 1;
int val = y;
//Loop by changing exponent in the powers of 2 and
//Find out low and high exponents between which the required exponent lies.
while(1)
{
val = pow((double)y, exp);
if(val == x)
return true;
else if(val > x)
break;
low = exp;
exp = exp * 2;
high = exp;
}
//Use binary search to find out the actual integer exponent if exists
//Otherwise, return false as no integer power.
int mid = (low + high)/2;
while(low < high)
{
val = pow((double)y, mid);
if(val > x)
{
high = mid-1;
}
else if(val == x)
{
return true;
}
else if(val < x)
{
low = mid+1;
}
mid = (low + high)/2;
}
return false;
}
int main()
{
cout<<isIntegerPower(1024,2);
}
double a=8;
double b=64;
double n = Math.log(b)/Math.log(a);
double e = Math.ceil(n);
if((n/e) == 1){
System.out.println("true");
} else{
System.out.println("false");
}
I would implement the function like so:
bool IsWholeNumberPower(int x, int y)
{
double power = log(x)/log(y);
return floor(power) == power;
}
This shouldn't need check within a delta as is common with floating point comparisons, since we're checking whole numbers.
On second thoughts, don't do this. It does not work for negative x and/or y. Note that all other log-based answers presented right now are also broken in exactly the same manner.
The following is a fast general solution (in Java):
static boolean isPow(int x, int y) {
int logyx = (int)(Math.log(x) / Math.log(y));
return pow(y, logyx) == x || pow(y, logyx + 1) == x;
}
Where pow() is an integer exponentiation function such as the following in Java:
static int pow(int a, int b) {
return (int)Math.pow(a, b);
}
(This works due to the following guarantee provided by Math.pow: "If both arguments are integers, then the result is exactly equal to the mathematical result of raising the first argument to the power of the second argument...")
The reason to go with logarithms instead of repeated division is performance: while log is slower than division, it is slower by a small fixed multiple. At the same time it does remove the need for a loop and therefore gives you a constant-time algorithm.
In cases where y is 2, there is a quick approach that avoids the need for a loop. This approach can be extended to cases where y is some larger power of 2.
If x is a power of 2, the binary representation of x has a single set bit. There is a fairly simple bit-fiddling algorithm for counting the bits in an integer in O(log n) time where n is the bit-width of an integer. Many processors also have specialised instructions that can handle this as a single operation, about as fast as (for example) an integer negation.
To extend the approach, though, first take a slightly different approach to checking for a single bit. First determine the position of the least significant bit. Again, there is a simple bit-fiddling algorithm, and many processors have fast specialised instructions.
If this bit is the only bit, then (1 << pos) == x. The advantage here is that if you're testing for a power of 4, you can test for pos % 2 == 0 (the single bit is at an even position). Testing for a power of any power of two, you can test for pos % (y >> 1) == 0.
In principle, you could do something similar for testing for powers of 3 and powers of powers of 3. The problem is that you'd need a machine that works in base 3, which is a tad unlikely. You can certainly test any value x to see if its representation in base y has a single non-zero digit, but you'd be doing more work that you're already doing. The above exploits the fact that computers work in binary.
Probably not worth doing in the real world, though.
Here is a Python version which puts together the ideas of #salva and #Axn and is modified to not generate any numbers greater than those given and uses only simple storage (read, "no lists") by repeatedly paring away at the number of interest:
def perfect_base(b, n):
"""Returns True if integer n can be expressed as b**e where
n is a positive integer, else False."""
assert b > 1 and n >= b and int(n) == n and int(b) == b
# parity check
if not b % 2:
if n % 2:
return False # b,n is even,odd
if b == 2:
return n & (n - 1) == 0
if not b & (b - 1) and n & (n - 1):
return False # b == 2**m but n != 2**M
elif not n % 2:
return False # b,n is odd,even
while n >= b:
d = b
while d <= n:
n, r = divmod(n, d)
if r:
return False
d *= d
return n == 1
Previous answers are correct, I liked Paul's answer the best. It's Simple and clean.
Here is the Java implementation of what he suggested:
public static boolean isPowerOfaNumber(int baseOrg, int powerOrg) {
double base = baseOrg;
double power = powerOrg;
while (base % power == 0)
base = base / power;
// return true if base is equal 1
return base == 1;
}
in the case the number is too large ... use log function to reduce time complexity:
import math
base = int(input("Enter the base number: "))
for i in range(base,int(input("Enter the end of range: "))+1):
if(math.log(i) / math.log(base) % 1 == 0 ):
print(i)
If you have access to the largest power of y, that can be fitted inside the required datatype, this is a really slick way of solving this problem.
Lets say, for our case, y == 3. So, we would need to check if x is a power of 3.
Given that we need to check if an integer x is a power of 3, let us start thinking about this problem in terms of what information is already at hand.
1162261467 is the largest power of 3 that can fit into an Java int.
1162261467 = 3^19 + 0
The given x can be expressed as [(a power of 3) + (some n)]. I think it is fairly elementary to be able to prove that if n is 0(which happens iff x is a power of 3), 1162261467 % x = 0.
So, to check if a given integer x is a power of three, check if x > 0 && 1162261467 % x == 0.
Generalizing. To check if a given integer x is a power of a given integer y, check if x > 0 && Y % x == 0: Y is the largest power of y that can fit into an integer datatype.
The general idea is that if A is some power of Y, A can be expressed as B/Ya, where a is some integer and A < B. It follows the exact same principle for A > B. The A = B case is elementary.
I found this Solution
//Check for If A can be expressed as power of two integers
int isPower(int A)
{
int i,a;
double p;
if(A==1)
return 1;
for(int a=1; a<=sqrt(A);++a )
{
p=log(A)/log(a);
if(p-int(p)<0.000000001)
return 1;
}
return 0;
}
binarycoder.org
I saw this question, and pop up this idea.
There exists a constant time (pretty fast) method for integers of limited size (e.g. 32-bit integers).
Note that for an integer N that is a power of 3 the following is true:
For any M <= N that is a power of 3, M divides N.
For any M <= N that is not a power 3, M does not divide N.
The biggest power of 3 that fits into 32 bits is 3486784401 (3^20). This gives the following code:
bool isPower3(std::uint32_t value) {
return value != 0 && 3486784401u % value == 0;
}
Similarly for signed 32 bits it is 1162261467 (3^19):
bool isPower3(std::int32_t value) {
return value > 0 && 1162261467 % value == 0;
}
In general the magic number is:
== pow(3, floor(log(MAX) / log(3)))
Careful with floating point rounding errors, use a math calculator like Wolfram Alpha to calculate the constant. For example for 2^63-1 (signed int64) both C++ and Java give 4052555153018976256, but the correct value is 4052555153018976267.
while (n % 3 == 0) {
n /= 3;
}
return n == 1;
Note that 1 is the zeroth power of three.
Edit: You also need to check for zero before the loop, as the loop will not terminate for n = 0 (thanks to Bruno Rothgiesser).
I find myself slightly thinking that if by 'integer' you mean 'signed 32-bit integer', then (pseudocode)
return (n == 1)
or (n == 3)
or (n == 9)
...
or (n == 1162261467)
has a certain beautiful simplicity to it (the last number is 3^19, so there aren't an absurd number of cases). Even for an unsigned 64-bit integer there still be only 41 cases (thanks #Alexandru for pointing out my brain-slip). And of course would be impossible for arbitrary-precision arithmetic...
I'm surprised at this. Everyone seems to have missed the fastest algorithm of all.
The following algorithm is faster on average - and dramatically faster in some cases - than a simple while(n%3==0) n/=3; loop:
bool IsPowerOfThree(uint n)
{
// Optimizing lines to handle the most common cases extremely quickly
if(n%3 != 0) return n==1;
if(n%9 != 0) return n==3;
// General algorithm - works for any uint
uint r;
n = Math.DivRem(n, 59049, out r); if(n!=0 && r!=0) return false;
n = Math.DivRem(n+r, 243, out r); if(n!=0 && r!=0) return false;
n = Math.DivRem(n+r, 27, out r); if(n!=0 && r!=0) return false;
n += r;
return n==1 || n==3 || n==9;
}
The numeric constants in the code are 3^10, 3^5, and 3^3.
Performance calculations
In modern CPUs, DivRem is a often single instruction that takes a one cycle. On others it expands to a div followed by a mul and an add, which would takes more like three cycles altogether. Each step of the general algorithm looks long but it actually consists only of: DivRem, cmp, cmove, cmp, cand, cjmp, add. There is a lot of parallelism available, so on a typical two-way superscalar processor each step will likely execute in about 4 clock cycles, giving a guaranteed worst-case execution time of about 25 clock cycles.
If input values are evenly distributed over the range of UInt32, here are the probabilities associated with this algorithm:
Return in or before the first optimizing line: 66% of the time
Return in or before the second optimizing line: 89% of the time
Return in or before the first general algorithm step: 99.998% of the time
Return in or before the second general algorithm step: 99.99998% of the time
Return in or before the third general algorithm step: 99.999997% of the time
This algorithm outperforms the simple while(n%3==0) n/=3 loop, which has the following probabilities:
Return in the first iteration: 66% of the time
Return in the first two iterations: 89% of the time
Return in the first three iterations: 97% of the time
Return in the first four iterations: 98.8% of the time
Return in the first five iterations: 99.6% of the time ... and so on to ...
Return in the first twelve iterations: 99.9998% of the time ... and beyond ...
What is perhaps even more important, this algorithm handles midsize and large powers of three (and multiples thereof) much more efficiently: In the worst case the simple algorithm will consume over 100 CPU cycles because it will loop 20 times (41 times for 64 bits). The algorithm I present here will never take more than about 25 cycles.
Extending to 64 bits
Extending the above algorithm to 64 bits is trivial - just add one more step. Here is a 64 bit version of the above algorithm optimized for processors without efficient 64 bit division:
bool IsPowerOfThree(ulong nL)
{
// General algorithm only
ulong rL;
nL = Math.DivRem(nL, 3486784401, out rL); if(nL!=0 && rL!=0) return false;
nL = Math.DivRem(nL+rL, 59049, out rL); if(nL!=0 && rL!=0) return false;
uint n = (uint)nL + (uint)rL;
n = Math.DivRem(n, 243, out r); if(n!=0 && r!=0) return false;
n = Math.DivRem(n+r, 27, out r); if(n!=0 && r!=0) return false;
n += r;
return n==1 || n==3 || n==9;
}
The new constant is 3^20. The optimization lines are omitted from the top of the method because under our assumption that 64 bit division is slow, they would actually slow things down.
Why this technique works
Say I want to know if "100000000000000000" is a power of 10. I might follow these steps:
I divide by 10^10 and get a quotient of 10000000 and a remainder of 0. These add to 10000000.
I divide by 10^5 and get a quotient of 100 and a remainder of 0. These add to 100.
I divide by 10^3 and get a quotient of 0 and a remainderof 100. These add to 100.
I divide by 10^2 and get a quotient of 1 and a remainder of 0. These add to 1.
Because I started with a power of 10, every time I divided by a power of 10 I ended up with either a zero quotient or a zero remainder. Had I started out with anything except a power of 10 I would have sooner or later ended up with a nonzero quotient or remainder.
In this example I selected exponents of 10, 5, and 3 to match the code provided previously, and added 2 just for the heck of it. Other exponents would also work: There is a simple algorithm for selecting the ideal exponents given your maximum input value and the maximum power of 10 allowed in the output, but this margin does not have enough room to contain it.
NOTE: You may have been thinking in base ten throughout this explanation, but the entire explanation above can be read and understood identically if you're thinking in in base three, except the exponents would have been expressed differently (instead of "10", "5", "3" and "2" I would have to say "101", "12", "10" and "2").
This is a summary of all good answers below this questions, and the performance figures can be found from the LeetCode article.
1. Loop Iteration
Time complexity O(log(n)), space complexity O(1)
public boolean isPowerOfThree(int n) {
if (n < 1) {
return false;
}
while (n % 3 == 0) {
n /= 3;
}
return n == 1;
}
2. Base Conversion
Convert the integer to a base 3 number, and check if it is written as a leading 1 followed by all 0. It is inspired by the solution to check if a number is power of 2 by doing n & (n - 1) == 0
Time complexity: O(log(n)) depending on language and compiler, space complexity: O(log(n))
public boolean isPowerOfThree(int n) {
return Integer.toString(n, 3).matches("^10*$");
}
3. Mathematics
If n = 3^i, then i = log(n) / log(3), and thus comes to the solution
Time complexity: depending on language and compiler, space complexity: O(1)
public boolean isPowerOfThree(int n) {
return (Math.log(n) / Math.log(3) + epsilon) % 1 <= 2 * epsilon;
}
4. Integer Limitations
Because 3^19 = 1162261467 is the largest power of 3 number fits in a 32 bit integer, thus we can do
Time complexity: O(1), space complexity: O(1)
public boolean isPowerOfThree(int n) {
return n > 0 && 1162261467 % n == 0;
}
5. Integer Limitations with Set
The idea is similar to #4 but use a set to store all possible power of 3 numbers (from 3^0 to 3^19). It makes code more readable.
6. Recursive (C++11)
This solution is specific to C++11, using template meta programming so that complier will replace the call isPowerOf3<Your Input>::cValue with calculated result.
Time complexity: O(1), space complexity: O(1)
template<int N>
struct isPowerOf3 {
static const bool cValue = (N % 3 == 0) && isPowerOf3<N / 3>::cValue;
};
template<>
struct isPowerOf3<0> {
static const bool cValue = false;
};
template<>
struct isPowerOf3<1> {
static const bool cValue = true;
};
int main() {
cout<<isPowerOf3<1162261467>::cValue;
return 0;
}
if (log n) / (log 3) is integral then n is a power of 3.
Recursively divide by 3, check that the remainder is zero and re-apply to the quotient.
Note that 1 is a valid answer as 3 to the zero power is 1 is an edge case to beware.
Very interesting question, I like the answer from starblue,
and this is a variation of his algorithm which will converge little bit faster to the solution:
private bool IsPow3(int n)
{
if (n == 0) return false;
while (n % 9 == 0)
{
n /= 9;
}
return (n == 1 || n == 3);
}
Between powers of two there is at most one power of three.
So the following is a fast test:
Find the binary logarithm of n by finding the position of the leading 1 bit in the number. This is very fast, as modern processors have a special instruction for that. (Otherwise you can do it by bit twiddling, see Bit Twiddling Hacks).
Look up the potential power of three in a table indexed by this position and compare to n (if there is no power of three you can store any number with a different binary logarithm).
If they are equal return yes, otherwise no.
The runtime depends mostly on the time needed for accessing the table entry. If we are using machine integers the table is small, and probably in cache (we are using it many millions of times, otherwise this level of optimization wouldn't make sense).
Here is a nice and fast implementation of Ray Burns' method in C:
bool is_power_of_3(unsigned x) {
if (x > 0x0000ffff)
x *= 0xb0cd1d99; // multiplicative inverse of 59049
if (x > 0x000000ff)
x *= 0xd2b3183b; // multiplicative inverse of 243
return x <= 243 && ((x * 0x71c5) & 0x5145) == 0x5145;
}
It uses the multiplicative inverse trick for to first divide by 3^10 and then by 3^5. Finally, it needs to check whether the result is 1, 3, 9, 27, 81, or 243, which is done by some simple hashing that I found by trial-and-error.
On my CPU (Intel Sandy Bridge), it is quite fast, but not as fast as the method of starblue that uses the binary logarithm (which is implemented in hardware on that CPU). But on a CPU without such an instruction, or when lookup tables are undesirable, it might be an alternative.
How large is your input? With O(log(N)) memory you can do faster, O(log(log(N)). Precompute the powers of 3 and then do a binary search on the precomputed values.
Simple and constant-time solution:
return n == power(3, round(log(n) / log(3)))
For really large numbers n, you can use the following math trick to speed up the operation of
n % 3 == 0
which is really slow and most likely the choke point of any algorithm that relies on repeated checking of remainders. You have to understand modular arithmetic to follow what I am doing, which is part of elementary number theory.
Let x = Σ k a k 2 k be the number of interest. We can let the upper bound of the sum be ∞ with the understanding that a k = 0 for some k > M. Then
0 ≡ x ≡ Σ k a k 2 k ≡ Σ k a 2k 2 2k + a 2k+1 2 2k+1 ≡ Σ k 2 2k ( a 2k + a 2k+1 2) ≡ Σ k a 2k + a 2k+1 2 (mod 3)
since 22k ≡ 4 k ≡ 1k ≡ 1 (mod 3).
Given a binary representation of a number x with 2n+1 bits as
x0 x1 x2 ... x2n+1
where xk ∈{0,1} you can group odd even pairs
(x0 x1) (x2 x3) ... (x2n x2n+1).
Let q denote the number of pairings of the form (1 0) and let r denote the number of pairings of the form (0 1). Then it follows from the equation above that 3 | x if and only if 3 | (q + 2r). Furthermore, you can show that 3|(q + 2r) if and only if q and r have the same remainder when divided by 3.
So an algorithm for determining whether a number is divisible by 3 could be done as follows
q = 0, r = 0
for i in {0,1, .., n}
pair <- (x_{2i} x_{2i+1})
if pair == (1 0)
switch(q)
case 0:
q = 1;
break;
case 1:
q = 2;
break;
case 2:
q = 0;
break;
else if pair == (0 1)
switch(r)
case 0:
r = 1;
break;
case 1:
r = 2;
break;
case 2:
r = 0;
return q == r
This algorithm is more efficient than the use of %.
--- Edit many years later ----
I took a few minutes to implement a rudimentary version of this in python that checks its true for all numbers up to 10^4. I include it below for reference. Obviously, to make use of this one would implement this as close to hardware as possible. This scanning technique can be extended to any number that one wants to by altering the derivation. I also conjecture the 'scanning' portion of the algorithm can be reformulated in a recursive O(log n) type formulation similar to a FFT, but I'd have to think on it.
#!/usr/bin/python
def bits2num(bits):
num = 0
for i,b in enumerate(bits):
num += int(b) << i
return num
def num2bits(num):
base = 0
bits = list()
while True:
op = 1 << base
if op > num:
break
bits.append(op&num !=0)
base += 1
return "".join(map(str,map(int,bits)))[::-1]
def div3(bits):
n = len(bits)
if n % 2 != 0:
bits = bits + '0'
n = len(bits)
assert n % 2 == 0
q = 0
r = 0
for i in range(n/2):
pair = bits[2*i:2*i+2]
if pair == '10':
if q == 0:
q = 1
elif q == 1:
q = 2
elif q == 2:
q = 0
elif pair == '01':
if r == 0:
r = 1
elif r == 1:
r = 2
elif r == 2:
r = 0
else:
pass
return q == r
for i in range(10000):
truth = (i % 3) == 0
bits = num2bits(i)
check = div3(bits)
assert truth == check
You can do better than repeated division, which takes O(lg(X) * |division|) time. Essentially you do a binary search on powers of 3. Really we will be doing a binary search on N, where 3^N = input value). Setting the Pth binary digit of N corresponds to multiplying by 3^(2^P), and values of the form 3^(2^P) can be computed by repeated squaring.
Algorithm
Let the input value be X.
Generate a list L of repeated squared values which ends once you pass X.
Let your candidate value be T, initialized to 1.
For each E in reversed L, if T*E <= X then let T *= E.
Return T == X.
Complexity:
O(lg(lg(X)) * |multiplication|)
- Generating and iterating over L takes lg(lg(X)) iterations, and multiplication is the most expensive operation in an iteration.
The fastest solution is either testing if n > 0 && 3**19 % n == 0 as given in another answer or perfect hashing (below). First I'm giving two multiplication-based solutions.
Multiplication
I wonder why everybody missed that multiplication is much faster than division:
for (int i=0, pow=1; i<=19, pow*=3; ++i) {
if (pow >= n) {
return pow == n;
}
}
return false;
Just try all powers, stop when it grew too big. Avoid overflow as 3**19 = 0x4546B3DB is the biggest power fitting in signed 32-bit int.
Multiplication with binary search
Binary search could look like
int pow = 1;
int next = pow * 6561; // 3**8
if (n >= next) pow = next;
next = pow * 81; // 3**4
if (n >= next) pow = next;
next = pow * 81; // 3**4; REPEATED
if (n >= next) pow = next;
next = pow * 9; // 3**2
if (n >= next) pow = next;
next = pow * 3; // 3**1
if (n >= next) pow = next;
return pow == next;
One step is repeated, so that the maximum exponent 19 = 8+4+4+2+1 can exactly be reached.
Perfect hashing
There are 20 powers of three fitting into a signed 32-bit int, so we take a table of 32 elements. With some experimentation, I found the perfect hash function
def hash(x):
return (x ^ (x>>1) ^ (x>>2)) & 31;
mapping each power to a distinct index between 0 and 31. The remaining stuff is trivial:
// Create a table and fill it with some power of three.
table = [1 for i in range(32)]
// Fill the buckets.
for n in range(20): table[hash(3**n)] = 3**n;
Now we have
table = [
1162261467, 1, 3, 729, 14348907, 1, 1, 1,
1, 1, 19683, 1, 2187, 81, 1594323, 9,
27, 43046721, 129140163, 1, 1, 531441, 243, 59049,
177147, 6561, 1, 4782969, 1, 1, 1, 387420489]
and can test very fast via
def isPowerOfThree(x):
return table[hash(x)] == x
Your question is fairly easy to answer by defining a simple function to run the check for you. The example implementation shown below is written in Python but should not be difficult to rewrite in other languages if needed. Unlike the last version of this answer, the code shown below is far more reliable.
Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 08:06:12) [MSC v.1900 64 bit (AMD64)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> import math
>>> def power_of(number, base):
return number == base ** round(math.log(number, base))
>>> base = 3
>>> for power in range(21):
number = base ** power
print(f'{number} is '
f'{"" if power_of(number, base) else "not "}'
f'a power of {base}.')
number += 1
print(f'{number} is '
f'{"" if power_of(number, base) else "not "}'
f'a power of {base}.')
print()
1 is a power of 3.
2 is not a power of 3.
3 is a power of 3.
4 is not a power of 3.
9 is a power of 3.
10 is not a power of 3.
27 is a power of 3.
28 is not a power of 3.
81 is a power of 3.
82 is not a power of 3.
243 is a power of 3.
244 is not a power of 3.
729 is a power of 3.
730 is not a power of 3.
2187 is a power of 3.
2188 is not a power of 3.
6561 is a power of 3.
6562 is not a power of 3.
19683 is a power of 3.
19684 is not a power of 3.
59049 is a power of 3.
59050 is not a power of 3.
177147 is a power of 3.
177148 is not a power of 3.
531441 is a power of 3.
531442 is not a power of 3.
1594323 is a power of 3.
1594324 is not a power of 3.
4782969 is a power of 3.
4782970 is not a power of 3.
14348907 is a power of 3.
14348908 is not a power of 3.
43046721 is a power of 3.
43046722 is not a power of 3.
129140163 is a power of 3.
129140164 is not a power of 3.
387420489 is a power of 3.
387420490 is not a power of 3.
1162261467 is a power of 3.
1162261468 is not a power of 3.
3486784401 is a power of 3.
3486784402 is not a power of 3.
>>>
NOTE: The last revision has caused this answer to become nearly the same as TMS' answer.
Set based solution...
DECLARE #LastExponent smallint, #SearchCase decimal(38,0)
SELECT
#LastExponent = 79, -- 38 for bigint
#SearchCase = 729
;WITH CTE AS
(
SELECT
POWER(CAST(3 AS decimal(38,0)), ROW_NUMBER() OVER (ORDER BY c1.object_id)) AS Result,
ROW_NUMBER() OVER (ORDER BY c1.object_id) AS Exponent
FROM
sys.columns c1, sys.columns c2
)
SELECT
Result, Exponent
FROM
CTE
WHERE
Exponent <= #LastExponent
AND
Result = #SearchCase
With SET STATISTICS TIME ON it record the lowest possible, 1 millisecond.
Another approach is to generate a table on compile time. The good thing is, that you can extend this to powers of 4, 5, 6, 7, whatever
template<std::size_t... Is>
struct seq
{ };
template<std::size_t N, std::size_t... Is>
struct gen_seq : gen_seq<N-1, N-1, Is...>
{ };
template<std::size_t... Is>
struct gen_seq<0, Is...> : seq<Is...>
{ };
template<std::size_t N>
struct PowersOfThreeTable
{
std::size_t indexes[N];
std::size_t values[N];
static constexpr std::size_t size = N;
};
template<typename LambdaType, std::size_t... Is>
constexpr PowersOfThreeTable<sizeof...(Is)>
generatePowersOfThreeTable(seq<Is...>, LambdaType evalFunc)
{
return { {Is...}, {evalFunc(Is)...} };
}
template<std::size_t N, typename LambdaType>
constexpr PowersOfThreeTable<N> generatePowersOfThreeTable(LambdaType evalFunc)
{
return generatePowersOfThreeTable(gen_seq<N>(), evalFunc);
}
template<std::size_t Base, std::size_t Exp>
struct Pow
{
static constexpr std::size_t val = Base * Pow<Base, Exp-1ULL>::val;
};
template<std::size_t Base>
struct Pow<Base, 0ULL>
{
static constexpr std::size_t val = 1ULL;
};
template<std::size_t Base>
struct Pow<Base, 1ULL>
{
static constexpr std::size_t val = Base;
};
constexpr std::size_t tableFiller(std::size_t val)
{
return Pow<3ULL, val>::val;
}
bool isPowerOfThree(std::size_t N)
{
static constexpr unsigned tableSize = 41; //choosen by fair dice roll
static constexpr PowersOfThreeTable<tableSize> table =
generatePowersOfThreeTable<tableSize>(tableFiller);
for(auto a : table.values)
if(a == N)
return true;
return false;
}
I measured times (C#, Platform target x64) for some solutions.
using System;
class Program
{
static void Main()
{
var sw = System.Diagnostics.Stopwatch.StartNew();
for (uint n = ~0u; n > 0; n--) ;
Console.WriteLine(sw.Elapsed); // nada 1.1 s
sw.Restart();
for (uint n = ~0u; n > 0; n--) isPow3a(n);
Console.WriteLine(sw.Elapsed); // 3^20 17.3 s
sw.Restart();
for (uint n = ~0u; n > 0; n--) isPow3b(n);
Console.WriteLine(sw.Elapsed); // % / 10.6 s
Console.Read();
}
static bool isPow3a(uint n) // Elric
{
return n > 0 && 3486784401 % n == 0;
}
static bool isPow3b(uint n) // starblue
{
if (n > 0) while (n % 3 == 0) n /= 3;
return n == 1;
}
}
Another way (of splitting hairs).
using System;
class Program
{
static void Main()
{
Random rand = new Random(0); uint[] r = new uint[512];
for (int i = 0; i < 512; i++)
r[i] = (uint)(rand.Next(1 << 30)) << 2 | (uint)(rand.Next(4));
var sw = System.Diagnostics.Stopwatch.StartNew();
for (int i = 1 << 23; i > 0; i--)
for (int j = 0; j < 512; j++) ;
Console.WriteLine(sw.Elapsed); // 0.3 s
sw.Restart();
for (int i = 1 << 23; i > 0; i--)
for (int j = 0; j < 512; j++) isPow3c(r[j]);
Console.WriteLine(sw.Elapsed); // 10.6 s
sw.Restart();
for (int i = 1 << 23; i > 0; i--)
for (int j = 0; j < 512; j++) isPow3b(r[j]);
Console.WriteLine(sw.Elapsed); // 9.0 s
Console.Read();
}
static bool isPow3c(uint n)
{ return (n & 1) > 0 && 3486784401 % n == 0; }
static bool isPow3b(uint n)
{ if (n > 0) while (n % 3 == 0) n /= 3; return n == 1; }
}
Python program to check whether the number is a POWER of 3 or not.
def power(Num1):
while Num1 % 3 == 0:
Num1 /= 3
return Num1 == 1
Num1 = int(input("Enter a Number: "))
print(power(Num1))
Python solution
from math import floor
from math import log
def IsPowerOf3(number):
p = int(floor(log(number) / log(3)))
power_floor = pow(3, p)
power_ceil = power_floor * 3
if power_floor == number or power_ceil == number:
return True
return False
This is much faster than the simple divide by 3 solution.
Proof: 3 ^ p = number
p log(3) = log(number) (taking log both side)
p = log(number) / log(3)
Here's a general algorithm for finding out if a number is a power of another number:
bool IsPowerOf(int n,int b)
{
if (n > 1)
{
while (n % b == 0)
{
n /= b;
}
}
return n == 1;
}
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
int main()
{
int n, power=0;
cout<<"enter a number"<<endl;
cin>>n;
if (n>0){
for(int i=0; i<=n; i++)
{
int r=n%3;
n=n/3;
if (r==0){
power++;
}
else{
cout<<"not exactly power of 3";
return 0;
}
}
}
cout<<"the power is "<<power<<endl;
}
This is a constant time method! Yes. O(1). For numbers of fixed length, say 32-bits.
Given that we need to check if an integer n is a power of 3, let us start thinking about this problem in terms of what information is already at hand.
1162261467 is the largest power of 3 that can fit into an Java int.
1162261467 = 3^19 + 0
The given n can be expressed as [(a power of 3) + (some x)]. I think it is fairly elementary to be able to prove that if x is 0(which happens iff n is a power of 3), 1162261467 % n = 0.
The general idea is that if X is some power of 3, X can be expressed as Y/3a, where a is some integer and X < Y. It follows the exact same principle for Y < X. The Y = X case is elementary.
So, to check if a given integer n is a power of three, check if n > 0 && 1162261467 % n == 0.
Python:
return n > 0 and 1162261467 % n == 0
OR Calculate log:
lg = round(log(n,3))
return 3**lg == n
1st approach is faster than the second one.
What is the best approach to calculating the largest prime factor of a number?
I'm thinking the most efficient would be the following:
Find lowest prime number that divides cleanly
Check if result of division is prime
If not, find next lowest
Go to 2.
I'm basing this assumption on it being easier to calculate the small prime factors. Is this about right? What other approaches should I look into?
Edit: I've now realised that my approach is futile if there are more than 2 prime factors in play, since step 2 fails when the result is a product of two other primes, therefore a recursive algorithm is needed.
Edit again: And now I've realised that this does still work, because the last found prime number has to be the highest one, therefore any further testing of the non-prime result from step 2 would result in a smaller prime.
Here's the best algorithm I know of (in Python)
def prime_factors(n):
"""Returns all the prime factors of a positive integer"""
factors = []
d = 2
while n > 1:
while n % d == 0:
factors.append(d)
n /= d
d = d + 1
return factors
pfs = prime_factors(1000)
largest_prime_factor = max(pfs) # The largest element in the prime factor list
The above method runs in O(n) in the worst case (when the input is a prime number).
EDIT:
Below is the O(sqrt(n)) version, as suggested in the comment. Here is the code, once more.
def prime_factors(n):
"""Returns all the prime factors of a positive integer"""
factors = []
d = 2
while n > 1:
while n % d == 0:
factors.append(d)
n /= d
d = d + 1
if d*d > n:
if n > 1: factors.append(n)
break
return factors
pfs = prime_factors(1000)
largest_prime_factor = max(pfs) # The largest element in the prime factor list
Actually there are several more efficient ways to find factors of big numbers (for smaller ones trial division works reasonably well).
One method which is very fast if the input number has two factors very close to its square root is known as Fermat factorisation. It makes use of the identity N = (a + b)(a - b) = a^2 - b^2 and is easy to understand and implement. Unfortunately it's not very fast in general.
The best known method for factoring numbers up to 100 digits long is the Quadratic sieve. As a bonus, part of the algorithm is easily done with parallel processing.
Yet another algorithm I've heard of is Pollard's Rho algorithm. It's not as efficient as the Quadratic Sieve in general but seems to be easier to implement.
Once you've decided on how to split a number into two factors, here is the fastest algorithm I can think of to find the largest prime factor of a number:
Create a priority queue which initially stores the number itself. Each iteration, you remove the highest number from the queue, and attempt to split it into two factors (not allowing 1 to be one of those factors, of course). If this step fails, the number is prime and you have your answer! Otherwise you add the two factors into the queue and repeat.
My answer is based on Triptych's, but improves a lot on it. It is based on the fact that beyond 2 and 3, all the prime numbers are of the form 6n-1 or 6n+1.
var largestPrimeFactor;
if(n mod 2 == 0)
{
largestPrimeFactor = 2;
n = n / 2 while(n mod 2 == 0);
}
if(n mod 3 == 0)
{
largestPrimeFactor = 3;
n = n / 3 while(n mod 3 == 0);
}
multOfSix = 6;
while(multOfSix - 1 <= n)
{
if(n mod (multOfSix - 1) == 0)
{
largestPrimeFactor = multOfSix - 1;
n = n / largestPrimeFactor while(n mod largestPrimeFactor == 0);
}
if(n mod (multOfSix + 1) == 0)
{
largestPrimeFactor = multOfSix + 1;
n = n / largestPrimeFactor while(n mod largestPrimeFactor == 0);
}
multOfSix += 6;
}
I recently wrote a blog article explaining how this algorithm works.
I would venture that a method in which there is no need for a test for primality (and no sieve construction) would run faster than one which does use those. If that is the case, this is probably the fastest algorithm here.
JavaScript code:
'option strict';
function largestPrimeFactor(val, divisor = 2) {
let square = (val) => Math.pow(val, 2);
while ((val % divisor) != 0 && square(divisor) <= val) {
divisor++;
}
return square(divisor) <= val
? largestPrimeFactor(val / divisor, divisor)
: val;
}
Usage Example:
let result = largestPrimeFactor(600851475143);
Here is an example of the code:
Similar to #Triptych answer but also different. In this example list or dictionary is not used. Code is written in Ruby
def largest_prime_factor(number)
i = 2
while number > 1
if number % i == 0
number /= i;
else
i += 1
end
end
return i
end
largest_prime_factor(600851475143)
# => 6857
All numbers can be expressed as the product of primes, eg:
102 = 2 x 3 x 17
712 = 2 x 2 x 2 x 89
You can find these by simply starting at 2 and simply continuing to divide until the result isn't a multiple of your number:
712 / 2 = 356 .. 356 / 2 = 178 .. 178 / 2 = 89 .. 89 / 89 = 1
using this method you don't have to actually calculate any primes: they'll all be primes, based on the fact that you've already factorised the number as much as possible with all preceding numbers.
number = 712;
currNum = number; // the value we'll actually be working with
for (currFactor in 2 .. number) {
while (currNum % currFactor == 0) {
// keep on dividing by this number until we can divide no more!
currNum = currNum / currFactor // reduce the currNum
}
if (currNum == 1) return currFactor; // once it hits 1, we're done.
}
//this method skips unnecessary trial divisions and makes
//trial division more feasible for finding large primes
public static void main(String[] args)
{
long n= 1000000000039L; //this is a large prime number
long i = 2L;
int test = 0;
while (n > 1)
{
while (n % i == 0)
{
n /= i;
}
i++;
if(i*i > n && n > 1)
{
System.out.println(n); //prints n if it's prime
test = 1;
break;
}
}
if (test == 0)
System.out.println(i-1); //prints n if it's the largest prime factor
}
The simplest solution is a pair of mutually recursive functions.
The first function generates all the prime numbers:
Start with a list of all natural numbers greater than 1.
Remove all numbers that are not prime. That is, numbers that have no prime factors (other than themselves). See below.
The second function returns the prime factors of a given number n in increasing order.
Take a list of all the primes (see above).
Remove all the numbers that are not factors of n.
The largest prime factor of n is the last number given by the second function.
This algorithm requires a lazy list or a language (or data structure) with call-by-need semantics.
For clarification, here is one (inefficient) implementation of the above in Haskell:
import Control.Monad
-- All the primes
primes = 2 : filter (ap (<=) (head . primeFactors)) [3,5..]
-- Gives the prime factors of its argument
primeFactors = factor primes
where factor [] n = []
factor xs#(p:ps) n =
if p*p > n then [n]
else let (d,r) = divMod n p in
if r == 0 then p : factor xs d
else factor ps n
-- Gives the largest prime factor of its argument
largestFactor = last . primeFactors
Making this faster is just a matter of being more clever about detecting which numbers are prime and/or factors of n, but the algorithm stays the same.
n = abs(number);
result = 1;
if (n mod 2 == 0) {
result = 2;
while (n mod 2 = 0) n /= 2;
}
for(i=3; i<sqrt(n); i+=2) {
if (n mod i == 0) {
result = i;
while (n mod i = 0) n /= i;
}
}
return max(n,result)
There are some modulo tests that are superflous, as n can never be divided by 6 if all factors 2 and 3 have been removed. You could only allow primes for i, which is shown in several other answers here.
You could actually intertwine the sieve of Eratosthenes here:
First create the list of integers up
to sqrt(n).
In the for loop mark all multiples
of i up to the new sqrt(n) as not
prime, and use a while loop instead.
set i to the next prime number in
the list.
Also see this question.
I'm aware this is not a fast solution. Posting as hopefully easier to understand slow solution.
public static long largestPrimeFactor(long n) {
// largest composite factor must be smaller than sqrt
long sqrt = (long)Math.ceil(Math.sqrt((double)n));
long largest = -1;
for(long i = 2; i <= sqrt; i++) {
if(n % i == 0) {
long test = largestPrimeFactor(n/i);
if(test > largest) {
largest = test;
}
}
}
if(largest != -1) {
return largest;
}
// number is prime
return n;
}
Python Iterative approach by removing all prime factors from the number
def primef(n):
if n <= 3:
return n
if n % 2 == 0:
return primef(n/2)
elif n % 3 ==0:
return primef(n/3)
else:
for i in range(5, int((n)**0.5) + 1, 6):
#print i
if n % i == 0:
return primef(n/i)
if n % (i + 2) == 0:
return primef(n/(i+2))
return n
I am using algorithm which continues dividing the number by it's current Prime Factor.
My Solution in python 3 :
def PrimeFactor(n):
m = n
while n%2==0:
n = n//2
if n == 1: # check if only 2 is largest Prime Factor
return 2
i = 3
sqrt = int(m**(0.5)) # loop till square root of number
last = 0 # to store last prime Factor i.e. Largest Prime Factor
while i <= sqrt :
while n%i == 0:
n = n//i # reduce the number by dividing it by it's Prime Factor
last = i
i+=2
if n> last: # the remaining number(n) is also Factor of number
return n
else:
return last
print(PrimeFactor(int(input())))
Input : 10
Output : 5
Input : 600851475143
Output : 6857
Inspired by your question I decided to implement my own version of factorization (and finding largest prime factor) in Python.
Probably the simplest to implement, yet quite efficient, factoring algorithm that I know is Pollard's Rho algorithm. It has a running time of O(N^(1/4)) at most which is much more faster than time of O(N^(1/2)) for trial division algorithm. Both algos have these running times only in case of composite (non-prime) number, that's why primality test should be used to filter out prime (non-factorable) numbers.
I used following algorithms in my code: Fermat Primality Test ..., Pollard's Rho Algorithm ..., Trial Division Algorithm. Fermat primality test is used before running Pollard's Rho in order to filter out prime numbers. Trial Division is used as a fallback because Pollard's Rho in very rare cases may fail to find a factor, especially for some small numbers.
Obviously after fully factorizing a number into sorted list of prime factors the largest prime factor will be the last element in this list. In general case (for any random number) I don't know of any other ways to find out largest prime factor besides fully factorizing a number.
As an example in my code I'm factoring first 190 fractional digits of Pi, code factorizes this number within 1 second, and shows largest prime factor which is 165 digits (545 bits) in size!
Try it online!
def is_fermat_probable_prime(n, *, trials = 32):
# https://en.wikipedia.org/wiki/Fermat_primality_test
import random
if n <= 16:
return n in (2, 3, 5, 7, 11, 13)
for i in range(trials):
if pow(random.randint(2, n - 2), n - 1, n) != 1:
return False
return True
def pollard_rho_factor(N, *, trials = 16):
# https://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm
import random, math
for j in range(trials):
i, stage, y, x = 0, 2, 1, random.randint(1, N - 2)
while True:
r = math.gcd(N, x - y)
if r != 1:
break
if i == stage:
y = x
stage <<= 1
x = (x * x + 1) % N
i += 1
if r != N:
return [r, N // r]
return [N] # Pollard-Rho failed
def trial_division_factor(n, *, limit = None):
# https://en.wikipedia.org/wiki/Trial_division
fs = []
while n & 1 == 0:
fs.append(2)
n >>= 1
d = 3
while d * d <= n and limit is None or d <= limit:
q, r = divmod(n, d)
if r == 0:
fs.append(d)
n = q
else:
d += 2
if n > 1:
fs.append(n)
return fs
def factor(n):
if n <= 1:
return []
if is_fermat_probable_prime(n):
return [n]
fs = trial_division_factor(n, limit = 1 << 12)
if len(fs) >= 2:
return sorted(fs[:-1] + factor(fs[-1]))
fs = pollard_rho_factor(n)
if len(fs) >= 2:
return sorted([e1 for e0 in fs for e1 in factor(e0)])
return trial_division_factor(n)
def demo():
import time, math
# http://www.math.com/tables/constants/pi.htm
# pi = 3.
# 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679
# 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196
# n = first 190 fractional digits of Pi
n = 1415926535_8979323846_2643383279_5028841971_6939937510_5820974944_5923078164_0628620899_8628034825_3421170679_8214808651_3282306647_0938446095_5058223172_5359408128_4811174502_8410270193_8521105559_6446229489
print('Number:', n)
tb = time.time()
fs = factor(n)
print('All Prime Factors:', fs)
print('Largest Prime Factor:', f'({math.log2(fs[-1]):.02f} bits, {len(str(fs[-1]))} digits)', fs[-1])
print('Time Elapsed:', round(time.time() - tb, 3), 'sec')
if __name__ == '__main__':
demo()
Output:
Number: 1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489
All Prime Factors: [3, 71, 1063541, 153422959, 332958319, 122356390229851897378935483485536580757336676443481705501726535578690975860555141829117483263572548187951860901335596150415443615382488933330968669408906073630300473]
Largest Prime Factor: (545.09 bits, 165 digits) 122356390229851897378935483485536580757336676443481705501726535578690975860555141829117483263572548187951860901335596150415443615382488933330968669408906073630300473
Time Elapsed: 0.593 sec
Here is my attempt in c#. The last print out is the largest prime factor of the number. I checked and it works.
namespace Problem_Prime
{
class Program
{
static void Main(string[] args)
{
/*
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
*/
long x = 600851475143;
long y = 2;
while (y < x)
{
if (x % y == 0)
{
// y is a factor of x, but is it prime
if (IsPrime(y))
{
Console.WriteLine(y);
}
x /= y;
}
y++;
}
Console.WriteLine(y);
Console.ReadLine();
}
static bool IsPrime(long number)
{
//check for evenness
if (number % 2 == 0)
{
if (number == 2)
{
return true;
}
return false;
}
//don't need to check past the square root
long max = (long)Math.Sqrt(number);
for (int i = 3; i <= max; i += 2)
{
if ((number % i) == 0)
{
return false;
}
}
return true;
}
}
}
#python implementation
import math
n = 600851475143
i = 2
factors=set([])
while i<math.sqrt(n):
while n%i==0:
n=n/i
factors.add(i)
i+=1
factors.add(n)
largest=max(factors)
print factors
print largest
Calculates the largest prime factor of a number using recursion in C++. The working of the code is explained below:
int getLargestPrime(int number) {
int factor = number; // assumes that the largest prime factor is the number itself
for (int i = 2; (i*i) <= number; i++) { // iterates to the square root of the number till it finds the first(smallest) factor
if (number % i == 0) { // checks if the current number(i) is a factor
factor = max(i, number / i); // stores the larger number among the factors
break; // breaks the loop on when a factor is found
}
}
if (factor == number) // base case of recursion
return number;
return getLargestPrime(factor); // recursively calls itself
}
Here is my approach to quickly calculate the largest prime factor.
It is based on fact that modified x does not contain non-prime factors. To achieve that, we divide x as soon as a factor is found. Then, the only thing left is to return the largest factor. It would be already prime.
The code (Haskell):
f max' x i | i > x = max'
| x `rem` i == 0 = f i (x `div` i) i -- Divide x by its factor
| otherwise = f max' x (i + 1) -- Check for the next possible factor
g x = f 2 x 2
The following C++ algorithm is not the best one, but it works for numbers under a billion and its pretty fast
#include <iostream>
using namespace std;
// ------ is_prime ------
// Determines if the integer accepted is prime or not
bool is_prime(int n){
int i,count=0;
if(n==1 || n==2)
return true;
if(n%2==0)
return false;
for(i=1;i<=n;i++){
if(n%i==0)
count++;
}
if(count==2)
return true;
else
return false;
}
// ------ nextPrime -------
// Finds and returns the next prime number
int nextPrime(int prime){
bool a = false;
while (a == false){
prime++;
if (is_prime(prime))
a = true;
}
return prime;
}
// ----- M A I N ------
int main(){
int value = 13195;
int prime = 2;
bool done = false;
while (done == false){
if (value%prime == 0){
value = value/prime;
if (is_prime(value)){
done = true;
}
} else {
prime = nextPrime(prime);
}
}
cout << "Largest prime factor: " << value << endl;
}
Found this solution on the web by "James Wang"
public static int getLargestPrime( int number) {
if (number <= 1) return -1;
for (int i = number - 1; i > 1; i--) {
if (number % i == 0) {
number = i;
}
}
return number;
}
Prime factor using sieve :
#include <bits/stdc++.h>
using namespace std;
#define N 10001
typedef long long ll;
bool visit[N];
vector<int> prime;
void sieve()
{
memset( visit , 0 , sizeof(visit));
for( int i=2;i<N;i++ )
{
if( visit[i] == 0)
{
prime.push_back(i);
for( int j=i*2; j<N; j=j+i )
{
visit[j] = 1;
}
}
}
}
void sol(long long n, vector<int>&prime)
{
ll ans = n;
for(int i=0; i<prime.size() || prime[i]>n; i++)
{
while(n%prime[i]==0)
{
n=n/prime[i];
ans = prime[i];
}
}
ans = max(ans, n);
cout<<ans<<endl;
}
int main()
{
ll tc, n;
sieve();
cin>>n;
sol(n, prime);
return 0;
}
Guess, there is no immediate way but performing a factorization, as examples above have done, i.e.
in a iteration you identify a "small" factor f of a number N, then continue with the reduced problem "find largest prime factor of N':=N/f with factor candidates >=f ".
From certain size of f the expected search time is less, if you do a primality test on reduced N', which in case confirms, that your N' is already the largest prime factor of initial N.
Here is my attempt in Clojure. Only walking the odds for prime? and the primes for prime factors ie. sieve. Using lazy sequences help producing the values just before they are needed.
(defn prime?
([n]
(let [oddNums (iterate #(+ % 2) 3)]
(prime? n (cons 2 oddNums))))
([n [i & is]]
(let [q (quot n i)
r (mod n i)]
(cond (< n 2) false
(zero? r) false
(> (* i i) n) true
:else (recur n is)))))
(def primes
(let [oddNums (iterate #(+ % 2) 3)]
(lazy-seq (cons 2 (filter prime? oddNums)))))
;; Sieve of Eratosthenes
(defn sieve
([n]
(sieve primes n))
([[i & is :as ps] n]
(let [q (quot n i)
r (mod n i)]
(cond (< n 2) nil
(zero? r) (lazy-seq (cons i (sieve ps q)))
(> (* i i) n) (when (> n 1) (lazy-seq [n]))
:else (recur is n)))))
(defn max-prime-factor [n]
(last (sieve n)))
Recursion in C
Algorithm could be
Check if n is a factor or t
Check if n is prime. If so, remember n
Increment n
Repeat until n > sqrt(t)
Here's an example of a (tail)recursive solution to the problem in C:
#include <stdio.h>
#include <stdbool.h>
bool is_factor(long int t, long int n){
return ( t%n == 0);
}
bool is_prime(long int n0, long int n1, bool acc){
if ( n1 * n1 > n0 || acc < 1 )
return acc;
else
return is_prime(n0, n1+2, acc && (n0%n1 != 0));
}
int gpf(long int t, long int n, long int acc){
if (n * n > t)
return acc;
if (is_factor(t, n)){
if (is_prime(n, 3, true))
return gpf(t, n+2, n);
else
return gpf(t, n+2, acc);
}
else
return gpf(t, n+2, acc);
}
int main(int argc, char ** argv){
printf("%d\n", gpf(600851475143, 3, 0));
return 0;
}
The solution is composed of three functions. One to test if the candidate is a factor, another to test if that factor is prime, and finally one to compose those two together.
Some key ideas here are:
1- Stopping the recursion at sqrt(600851475143)
2- Only test odd numbers for factorness
3- Only testing candidate factors for primeness with odd numbers
It seems to me that step #2 of the algorithm given isn't going to be all that efficient an approach. You have no reasonable expectation that it is prime.
Also, the previous answer suggesting the Sieve of Eratosthenes is utterly wrong. I just wrote two programs to factor 123456789. One was based on the Sieve, one was based on the following:
1) Test = 2
2) Current = Number to test
3) If Current Mod Test = 0 then
3a) Current = Current Div Test
3b) Largest = Test
3c) Goto 3.
4) Inc(Test)
5) If Current < Test goto 4
6) Return Largest
This version was 90x faster than the Sieve.
The thing is, on modern processors the type of operation matters far less than the number of operations, not to mention that the algorithm above can run in cache, the Sieve can't. The Sieve uses a lot of operations striking out all the composite numbers.
Note, also, that my dividing out factors as they are identified reduces the space that must be tested.
Compute a list storing prime numbers first, e.g. 2 3 5 7 11 13 ...
Every time you prime factorize a number, use implementation by Triptych but iterating this list of prime numbers rather than natural integers.
With Java:
For int values:
public static int[] primeFactors(int value) {
int[] a = new int[31];
int i = 0, j;
int num = value;
while (num % 2 == 0) {
a[i++] = 2;
num /= 2;
}
j = 3;
while (j <= Math.sqrt(num) + 1) {
if (num % j == 0) {
a[i++] = j;
num /= j;
} else {
j += 2;
}
}
if (num > 1) {
a[i++] = num;
}
int[] b = Arrays.copyOf(a, i);
return b;
}
For long values:
static long[] getFactors(long value) {
long[] a = new long[63];
int i = 0;
long num = value;
while (num % 2 == 0) {
a[i++] = 2;
num /= 2;
}
long j = 3;
while (j <= Math.sqrt(num) + 1) {
if (num % j == 0) {
a[i++] = j;
num /= j;
} else {
j += 2;
}
}
if (num > 1) {
a[i++] = num;
}
long[] b = Arrays.copyOf(a, i);
return b;
}
This is probably not always faster but more optimistic about that you find a big prime divisor:
N is your number
If it is prime then return(N)
Calculate primes up until Sqrt(N)
Go through the primes in descending order (largest first)
If N is divisible by Prime then Return(Prime)
Edit: In step 3 you can use the Sieve of Eratosthenes or Sieve of Atkins or whatever you like, but by itself the sieve won't find you the biggest prime factor. (Thats why I wouldn't choose SQLMenace's post as an official answer...)
Here is the same function#Triptych provided as a generator, which has also been simplified slightly.
def primes(n):
d = 2
while (n > 1):
while (n%d==0):
yield d
n /= d
d += 1
the max prime can then be found using:
n= 373764623
max(primes(n))
and a list of factors found using:
list(primes(n))
I think it would be good to store somewhere all possible primes smaller then n and just iterate through them to find the biggest divisior. You can get primes from prime-numbers.org.
Of course I assume that your number isn't too big :)
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include <time.h>
factor(long int n)
{
long int i,j;
while(n>=4)
{
if(n%2==0) { n=n/2; i=2; }
else
{ i=3;
j=0;
while(j==0)
{
if(n%i==0)
{j=1;
n=n/i;
}
i=i+2;
}
i-=2;
}
}
return i;
}
void main()
{
clock_t start = clock();
long int n,sp;
clrscr();
printf("enter value of n");
scanf("%ld",&n);
sp=factor(n);
printf("largest prime factor is %ld",sp);
printf("Time elapsed: %f\n", ((double)clock() - start) / CLOCKS_PER_SEC);
getch();
}